Can the diameter be controled by the injectivity radius and the volume?

Question

Diameter bounded from above is usually needed in the finiteness theorem or other convergence theorems in Riemannian Geometry. Let $M^n$ be a closed manifold and {$g_i$} be a family of smooth Riemannian metrics on it with $Inj_{g_i}\geq \alpha>0$ and $Vol_{g_i}\leq \beta$. Are those two conditions enough to imply $Diam_{g_i}\leq \gamma$? If so, does the $\gamma$ depend only on $\alpha,\beta$ and $n$?

Answer 1

By Croke, Some isoperimetric inequalities and eigenvalue estimates, Proposition 14, on an $n$-dimensional Riemannian manifold with injectivity radius $\alpha$, a ball of radius $\alpha/2$ has volume at least $C\alpha^n/ (2n)^n$ for a constant $C$. Specifically, by their theorem 11, $C=2^{n-1}V_{n-1}^{n\phantom{1}} / V_{n\phantom{1}}^{n-1}$, where $V_n$ is the volume of the $n$-dimensional sphere.

Let $x_1,\dots,x_m$ be a maximal collection of points such that the balls of radius $\alpha/2$ centered at $x_1,\dots,x_m$ do not overlap. Then clearly $$m\frac{ C\alpha^n }{ (2n)^n} \leq \beta.$$ By maximality, the balls of radius $\alpha$ centered at $x_1,\dots, x_m$ cover $M$. Hence because $M^n$ is connected, the graph with vertices $x_1,\dots,x_m$ and edges connecting $x_i$ to $x_j$ if the distance from $x_i$ to $x_j$ is at most $2\alpha$ is connected and thus has diameter at most $m-1$.

Thus the diameter is at most $$\alpha + (m-1) 2\alpha+ \alpha = 2m\alpha \leq 2 \beta \frac{ (2n)^n}{ C \alpha^n} \alpha = \frac{ \beta}{\alpha^{n-1} } \frac{2^{n+1} n^n} {C} $$ as for the distance between $x$ and $y$, it takes a distance of $\alpha$ to get from $x$ to one of the $x_i$, then $(m-1)$ distances of $2\alpha$ to travel to an appropriate $x_j$, and then $\alpha$ to get from $x_j$ to $y$.

This is sharp up to a constant depending on $n$, as demonstrated by a suitable long thin torus.